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HW05_2011 solution

# HW05_2011 solution - Problem Set 5 Solutions(1(a Pr Y& y...

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Unformatted text preview: Problem Set 5 Solutions (1) (a) Pr( Y & y ) = Pr( j X j 3 & y ) = Pr( j X j & 3 p y ) = Pr( Â¡ 3 p y & X & 3 p y ) = Pr( X & 3 p y ) Â¡ Pr( X & Â¡ 3 p y ) = F X ( 3 p y ) Â¡ F X ( Â¡ 3 p y ) f Y ( y ) = @F X ( 3 p y ) @y Â¡ @F X ( & 3 p y ) @y = f X ( 3 p y ) 1 3 y & 2 3 + f X ( Â¡ 3 p y ) 1 3 y & 2 3 = 1 2 e & 3 p y 1 3 y & 2 3 + 1 2 e & 3 p y 1 3 y & 2 3 = 1 3 e & 3 p y y & 2 3 ; < y < 1 : (b) Pr( Y & y ) = Pr(1 Â¡ X 2 & y ) = Pr( X 2 Â¢ 1 Â¡ y ) = Pr( j X j Â¢ p 1 Â¡ y ) = Pr( X Â¢ p 1 Â¡ y ) + Pr( Â¡ X Â¢ p 1 Â¡ y ) = 1 Â¡ Pr( X & p 1 Â¡ y ) + Pr( X & Â¡ p 1 Â¡ y ) = 1 Â¡ F X ( p 1 Â¡ y ) + F X ( Â¡ p 1 Â¡ y ) f Y ( y ) = @F X ( p 1 & y ) @y + @F X ( & p 1 & y ) @y = Â¡ 3 8 (1 + p 1 Â¡ y ) 2 & 1 2 p 1 & y + 3 8 (1 + p 1 Â¡ y ) 2 1 2 p 1 & y = 3 8 2 & y p 1 & y ; < y < 4 (2) P ( Y & y ) = P ( X 2 & y ) = 8 > < > : P ( Â¡ p y < X < p y ) ; if j x j < 1 P (1 < X < p y ) ; if x Â¢ 1 = 8 > < > : R p y & p y f X ( x ) dx; if j x j < 1 R p y 1 f X ( x ) dx; if x Â¢ 1 Di/erentiation gives f Y ( y ) = 2 9 1 p y ; if y < 1 1 9 + 1 9 1 p y ; if y Â¢ 1 (3) Use the Probability Integral Transform theorem, u ( x ) = F X ( x ) = 8 > > > > > < > > > > > : , if x & 1 ( x Â¡ 1) 2 = 4 , if 1 <...
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